Let $a$ and $b$ be nonnegative real numbers such that
\[\sin (ax + b) = \sin 29x\]for all integers $x.$  Find the smallest possible value of $a.$
First, let $a$ and $b$ be nonnegative real numbers such that
\[\sin (ax + b) = \sin 29x\]for all integers $x.$  Let $a' = a + 2 \pi n$ for some integer $n.$  Then
\begin{align*}
\sin (a' x + b) &= \sin ((a + 2 \pi n) x + b) \\
&= \sin (ax + b + 2 \pi n x) \\
&= \sin (ax + b) \\
&= \sin 29x
\end{align*}for all integers $x.$

Conversely, suppose $a,$ $a',$ and $b$ are nonnegative real numbers such that
\[\sin (ax + b) = \sin (a'x + b) = \sin 29x \quad (*)\]for all integers $x.$  Then from the angle addition formula,
\[\sin ax \cos b + \cos ax \sin b = \sin a'x \cos b + \cos a'x \sin b = \sin 29x.\]Taking $x = 0$ in $(*),$ we get $\sin b = 0.$  Hence,
\[\sin ax \cos b = \sin a'x \cos b.\]Since $\cos b \neq 0,$
\[\sin ax = \sin a'x\]for all integers $x.$

Taking $x = 1,$ we get $\sin a = \sin a'.$  Taking $x = 2,$ we get $\sin 2a = \sin 2a'.$  From the angle addition formula,
\[\sin 2a = \sin a \cos a + \cos a \sin a = 2 \sin a \cos a.\]Similarly, $\sin 2a' = 2 \sin a' \cos a',$ so
\[2 \sin a \cos a = 2 \sin a' \cos a'.\]Taking $x = 1$ in $\sin ax \cos b = \sin a'x \cos b = \sin 29x,$ we get
\[\sin a \cos b = \sin a' \cos b = \sin 29,\]which means $\sin a = \sin a' \neq 0.$  Thus, we can safely divide both sides of $2 \sin a \cos a = 2 \sin a' \cos a'$ by $2 \sin a = 2 \sin a',$ to get
\[\cos a = \cos a'.\]Finally, since $\sin a = \sin a'$ and $\cos a = \cos a',$ $a$ and $a'$ must differ by a multiple of $2 \pi.$

In our work, we derived that if
\[\sin (ax + b) = \sin 29x\]for all integers $x,$ then $\sin b = 0,$ so $b$ is a multiple of $\pi.$  Since the sine function has period $2 \pi,$ we only need to consider the cases where $b = 0$ or $b = \pi.$

If $b = 0,$ then
\[\sin ax = \sin 29x\]for all integers $x.$  We see that $a = 29$ works, so the only solutions are of the form $a = 29 + 2k \pi,$ where $k$ is an integer.  The smallest nonnegative real number of this form is $a = 29 - 8 \pi.$

If $b = \pi,$ then
\[\sin (ax + \pi) = \sin 29x\]for all integers $x.$  We see that $a = -29$ works, since
\[\sin (-29x + \pi) = \sin (-29x) \cos \pi = \sin 29x.\]So the only solutions are of the form $a = -29 + 2k \pi,$ where $k$ is an integer.  The smallest nonnegative real number of this form is $a = -29 + 10 \pi.$

Thus, the smallest such constant $a$ is $\boxed{10 \pi - 29}.$